It's not a violation, because the decay itself is a current. Remeber the continuity equation, which is closely tied to the conservation of charge: \[\vec \nabla \cdot \vec J = -\frac{\partial \rho}{\partial t}\]

It's honestly more of a conservation of charge concept. Think about a capacitor. Even though current doesn't flow through, charge is still conserved as it collects on the plates, and thus we don't think twice about it when we apply Kirchoff's Current Law. If we think of each of the plates themselves as junctions, we find that the current flowing in is NOT the current flowing out. It's a similar concept here. It's not that it's a violation, it's rather that Kirchoff's Current Law has no place in microscopic electrodynamics.

How can it? Charge is always conserved. In beta decay you emit an electron, and the nucleus acquires another positive charge. That either leaves you with a cation -- in which case you have a current of both electrons and positive anions, i.e. a neutral current -- or the cation absorbs a nearby electron, which cancels out the charge of the beta particle.
It's true when you think about nuclear physics, you tend to ignore the behaviour of any orbital electrons, or indeed any electrons at thermal (and not nuclear) energies. But they're there, and in cases interact to balance the books.